help measuring true position

[[Me...]]

i am trying to measure true position on this part but my true position dimension is way off. my alignment is the cylinder of the part, top face of part. How do i approach this so it gives me the .0866 dimension i am trying to achieve? thank you.

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[[SH...]]

Use only one datum in your DRF, in the drawing only one datum is called.

[[Lo...]]

The TP callout on the print is not a diametrical tolerance zone. Looks like you should be evaluating in one direction for the TP callout.

[[Me...]]

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i removed the other datums but it still shows my dimension at .042(1.067mm).

[[Ja...]]

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This would be an "Only Y" tolerance zone.

[[Am...]]

As others have said it should be only in Y and also only use the A datum as shown on the print, in your pictures you are using a secondary and tertiary datum that is not only unnecessary it changes what you are reporting.

[[Ri...]]

Fill in Datum A for all three Datums and see what kind of results you get.

Once again, Calypso really like a fully constrained DRF, and in the case of most True Positions, it will just pull features from the Base Alignment to fully constrain it.

[[SH...]]

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No, using one feature we can't constrain all the degree of freedom.

[[Ri...]]

Your Position only have one Datum. It isn't fully constrained.

[[Br...]]

So many problems with this drawing. First, the Position tolerance isn't applied to a feature of size. If its supposed to be applied to that Hole in the side of the part, then the feature control frame needs to be placed under the Ø.1969 dimension. Then the .0866 dimension needs to be basic. Its a complete contradiction to have a tolerance on that dimension and simultaneously apply a position tolerance that governs that same dimension. You essentially have a ±0.04mm tolerance and a ±0.02mm tolerance on the same dimension. So which is it?

You also have to then specify what direction the position tolerance applies to if it is non cylindrical. And once again it must be applied to the feature of size, not the dimension as shown here. We can make a bunch of assumptions as to what this means, but since there are already tolerances placed on the dimensions, and the tolerance did not get placed on an actual feature of size- it's quite difficult to infer what the designer actually intends here without some other information not on this drawing.

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Richard is correct. The datum reference frame only constrains 4 degrees of freedom. You're free to translate and rotate on the other 2 degrees of freedom to optimize the feature's position within the tolerance zone.